### Question

Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.

Calling `next()`

will return the next smallest number in the BST.

**Note: **`next()`

and `hasNext()`

should run in average O(1) time and uses O(*h*) memory, where *h* is the height of the tree.

**Credits:**

Special thanks to @ts for adding this problem and creating all test cases.

### Analysis

This is an extremely important question, if you are going for an interview. I repeat: **this is an extremely important question, if you are going for an interview**. If you do not remember it by heart, I will repeat again.

The solution of the iterator applies to a lot of related questions. So make sure you practise this question until you are perfect. You WILL BE ASKED this question at times.

You could read my other post [Question] Iterator of Binary Search Tree.

### Solution

We only need to keep 1 variable in RAM, that is a stack.

### Code

```
public class BSTIterator {
Stack<TreeNode> stack = new Stack<TreeNode>();
public BSTIterator(TreeNode root) {
while (root != null) {
stack.push(root);
root = root.left;
}
}
/** @return whether we have a next smallest number */
public boolean hasNext() {
return !stack.isEmpty();
}
/** @return the next smallest number */
public int next() {
if (!hasNext()) {
return 0;
}
TreeNode next = stack.pop();
TreeNode node = next.right;
while (node != null) {
stack.push(node);
node = node.left;
}
return next.val;
}
}
```